I am looking at a simple example of choosing a point at random on a disk (circle) of radius $1$. The text book states simply:
$f(x,y) = \frac{1}{\pi}$ if $x^2 + y^2 \leq 1$ and $0$ otherwise. Now, when thinking in terms of single variable integrals, I understand that to find the pdf of a uniform random variable $X \in [a,b]$ we simply have the pdf equal to $\frac{1}{b-a}$ and this function integrates to $1$ over the range $[a,b]$. I know a similar version is at work here in the multivariate case but can't as easily see the justification or how to set-up the double integral to integrate over the circle using the pdf $\frac{1}{\pi}$ to get $1$.
I assume that for any area $A$ we simply have $\frac{1}{A}$ as our pdf. Can someone give me the generic double integral that we would use for an arbitrary area $A$? I suppose for a square or rectangle it seems straightforward $[a,b]\times[c,d]$ would imply a double integral of the form$\int_{a}^{b} \int_{c}^{d}$ but for a circle or a more complex area I am a bit lost.
Can someone shine some light on ways to think about the multivariate case of deriving the pdf and also provide guidance on how to set-up the double integral that would integrate $\frac{1}{\pi}$ to $1$? I am having a hard time figuring out the limits of integration.
To integrate over a unit circle, $y$ should range from $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$ while $x$ stays inside $[-1,1]$ (look at the circle to see this). This is a slightly annoying integral to do directly, but can be done using trig substitutions:
$$ \int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\dfrac{1}{\pi}dydx = 1$$
Otherwise, it'll be easier to convert to polar coordinates:
$$ \int_0^{2\pi}\int_0^1\dfrac{1}{\pi}rdrd\theta = 1$$